Two Counterexamples Concerning the Pluri-Complex Green Function in Cn
Eric Bedford and Jean-Pierre Demailly*
Indiana University and Universit´e de Grenoble I
Given a domain Ω⊂Cn and a pointz ∈Ω, we consider the function uz(ζ) = sup{v(ζ) :v is psh on Ω, v <0, and v(ζ)6log|ζ−z|+O(1)}, which is the pluri-complex Green function on Ω with logarithmic pole at z (cf. [2,3,4]). For a large class of domains Ω, (e.g. if∂Ω is strongly pseudoconvex and bounded), then uz ∈C∞( ¯Ω− {z}), uz = 0 on∂Ω, anduz(ζ) = log|z−ζ|+O(1).
Several further properties of uz were derived in [2], where the following questions were also raised:
1. Is uz ∈C2( ¯Ω− {z}) ?
2. Is uz symmetric, i.e. uz(ζ) =uζ(z) ?
Lempert [4] has shown that if Ω ⊂ Cn is strictly convex and smoothly bounded, then the answer to these questions is “Yes.” The convex situation, however, is quite special, and the point of this note is to show that for strongly pseudoconvex domains the answer to both these questions is “No.”
Proposition 1. — Let Ω ⊂ C2 be a bounded, connected, strongly pseudo- convex domain with C2 boundary. Let Ω be invariant under the transformation (z, w)7→(z,−w), and suppose thatΩ∩{w= 0}=D1∪. . .∪Ds, whereD¯i∩D¯j =∅, and Dj is a smoothly bounded disk. Then if p∈D1, the function up is not C2 in a neighborhood of the set D¯j for any 26j 6s.
Remark. — For a specific example, we may let Ω be a small neighborhood of the unit circle {(x1, x2)∈R2 ⊂C2 :x21+x22 = 1}.
Proof. — Since both pand the domain Ω are invariant under the mapping (z, w)7→(z,−w), it follows thatup is also invariant. Ifup is of class C2 on a neigh- borhood of the set ¯D2, then we have an invariant solution of det(∂2up/∂zi∂z¯j) = 0 on ¯D2, with up = 0 on ∂D2. But this is a contradiction to the Proposition of [1], which completes the proof.
Proposition 2. — There is a bounded, strongly pseudoconvex domain Ωin C2 with real analytic boundary for which the Green function is not symmetric.
* This paper was written while the authors were visiting the Mittag-Leffler Institute, and they wish to thank the Institute for its hospitality.
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Proof. — Let u(z, w) = max 12 log(|w2−z2(z−a)|/ε2),log|z|
, where
|a| < 1. We will show that for ε > 0 small, the Green function for the domain Ω = {u(z, w) < 0}, is not symmetric. To prove Proposition 2, we consider an increasing sequence of strongly pseudoconvex domains Ωj with real analytic boundary, which increase to Ω. If the Green function uz(ζ,Ωj) is symmetric for all j, then so is limj→∞uz(ζ,Ωj) = uz(ζ). Thus one of the sets Ωj must give us the set desired in Proposition 2.
We note that u is the Green function for the domain Ω with pole at (0,0).
Thus u(0,0)(a,0) = u(a,0) = log|a|. On the other hand, the pole of u(a,0) is estimated by the function
v(z, w) =
log|w|/2 if |z−a|6ε ,
max log|w|/2, 1
2 + 1
plog 1/ε
! log
z −a 1−¯az
!
if |z−a|>ε ,
Indeed, we have |w|<2 on Ω, and thus v <0. Moreover the inequalities
|w|2 >|z2(z−a)| −ε2 >(|a| −ε)ε−ε2
on Ω∩ {|z−a|=ε}show thatv is psh. As v(z, w)6log(|w|+|z−a|) +O(1) near (a,0), it follows that v(z, w)6u(a,0)(z, w). Thus if ε is small, we have
u(a,0)(0,0)> 1
2 + 1
plog 1/ε
!
log|a|> u(0,0)(a,0),
and so the Green function is not symmetric.
Remark. In general, the Green function uz(ζ) cannot even be expected to be psh in z. In fact, for any domain Ω, the symmetry property is equivalent to the fact that v(z) = uz(ζ) is psh in z for every fixed ζ. To see this, observe that v < 0 and v(z) = log|z−ζ|+O(1) near ζ. the plurisubharmonicity of v implies therefore uz(ζ) =v(z)6uζ(z), and similarly uζ(z)6uz(ζ).
References
[1] E. Bedford and J.-E. Fornæss. — Counterexamples to regularity for the complex Monge-Amp`ere equation, Inv. Math.,50(), 129–134.
[2] J.-P. Demailly. — Mesures de Monge-Amp`ere et mesures pluriharmoniques, Math.
Z.,194 (), 519–564.
[3] M. Klimek. — Extremal plurisubharmonic functions and invariant pseudodistances, Bull. Soc. Math. France, 113(), 123–142.
[4] L. Lempert. — La m´etrique de Kobayashi et la repr´esentation des domaines sur la boule, Bull. Soc. Math. France,109 (), 427–474.
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